Answer: The distance is 723.4km
Explanation:
The velocity of the transverse waves is 8.9km/s
The velocity of the longitudinal wave is 5.1 km/s
The transverse one reaches 68 seconds before the longitudinal.
if the distance is X, we know that:
X/(9.8km/s) = T1
X/(5.1km/s) = T2
T2 = T1 + 68s
Where T1 and T2 are the time that each wave needs to reach the sesmograph.
We replace the third equation into the second and get:
X/(9.8km/s) = T1
X/(5.1km/s) = T1 + 68s
Now, we can replace T1 from the first equation into the second one:
X/(5.1km/s) = X/(9.8km/s) + 68s
Now we can solve it for X and find the distance.
X/(5.1km/s) - X/(9.8km/s) = 68s
X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s
X = 68s/0.094s/km = 723.4 km
terminal velocity ... greater speed ... acc is 10m/s/s
Answer
D. 0.25 meters/second2
Explanation
The average acceleration is the ratio of change in velocity to the change in time of travel.Taking in this case that the change of velocity is a unit, then Average acceleration is given by;
Aacc=Vf-Vi/Tf-Ti
where Vf=final velocity,Vi=initial velocity' Tf=final time, Ti=initial time
Vf-Vi=1m/s
Tf-Ti=4-0=4seconds
Avacc=1/4=0.25m/s2
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